Ordinary differential equation
.]] An ordinary differential equation or ODE (as opposed to a partial differential equation) is a type of differential equation that involves a function of only one independent variable. It can most simply be defined, for a layman, as any equation that involves any combination of the following: * An independent variable ( x ) * Functions of the independent variable (or dependent variables) ( g(x) ) * A primary dependent variable (the function in question) ( y(x) ) * And, necessarily, any number of and degrees of derivatives of the primary function ( \frac{dy}{dx}, \frac{d^2 y}{dx^2} ) An example differential equation is as follows: x^2 y + 3 \frac{d^2 y}{dx^2} - xy\frac{d^2 y}{dx^2} = 3 Methods of solving First order First order first degree differential equations in the form : \frac{dy}{dx} = f(x) Can be solved with direct integration. Separable equations Separable equations are the one of the easiest to solve. For any ODE in the form : \frac{dy}{dx} = f(x) g(y) the solution is : \int \frac{dy}{g(y)} = \int f(x) dx General linear equations For linear equations in the form : \frac{dy}{dx} + f(x)y = g(x) , the solution can be found with the formula : y = \frac{1}{\mu} \int \mu g(x) dx, \mu = e^{\int f(x)dx} Bernoulli equations Bernoulli differential equations are those in the form : \frac{dy}{dx} + f(x)y = g(x)y^n , They can be solved by transforming them into linear differential equations by substituting. Exact equations Exact differential equations are those in the form : m(x,y) dx + n(x,y) dy = 0 \, \, \text{if} \, \, \frac{\partial }{\partial y} m(x,y) = \frac{\partial}{\partial x} n(x,y) and will have the solution : \int m(x,y)dx + \int n(x,y)dy = C Systems of differential equations , one can create a vector field called the phase portrait based on the mapping from x'' (the vector representing a given point) to ''x', similarly to a slope field. This particular phase portrait is based on the equation \tfrac{d^2y}{dt^2}+(y^2-1)\tfrac{dy}{dt}+y=0. ]] A system of ODEs in the form : \begin{bmatrix} x'_1 \\ \vdots \\ x'_n \end{bmatrix} = \begin{bmatrix} C_{11} & \cdots & C_{1n} \\ \vdots & \ddots & \vdots\\ C_{n1} & \cdots & C_{nn} \end{bmatrix} \begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix} or, in more compact notation, : \vec{x}' = A \vec{x} the general solution will be : \vec{x} = C_1 \vec{v}_1 e^{\lambda_1 t} + \cdots + C_n \vec{v}_n e^{\lambda_n t} where \vec{v} and \lambda represent the eigenvectors and eigenvalues of A'', respectively. Higher order differential equations can be converted into such a system by making the substitution : x_n = y^{(n)} (t) differentiating, and substituting variables from the original equation for derivatives of ''y to yield a system of first order ODEs. For example: : y'' - 2y' + y = 0 We can make the substition : \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} y \\ y' \end{bmatrix} : \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}' = \begin{bmatrix} y' \\ y'' \end{bmatrix} = \begin{bmatrix} x_2 \\ 2x_2 - x_1 \end{bmatrix} : \vec{x}' = \begin{bmatrix} 0 & 1 \\ 2 & -1 \end{bmatrix} \vec{x} Higher order equations In general, if f(x) and g(x) are both solutions to a differential equation, C1f(x) + C2g(x) is also a solution. Homogenous equations : a\frac{d^2 y}{dx^2} + b\frac{dy}{dx}+ cy = 0 Differential equations in this form can be solved by first finding the roots of the auxiliary or characteristic equation, which is equal to : ar^2 + br + c = 0 The roots may be real or complex. If the roots are r_1, r_2 then the general solution will be: : y=C_1 e^{r_1 x} + C_2 e^{r_2 x} If the characteristic equation has only one solution, the general solution of the ODE will be : y=C_1 e^{r x} + C_2 x e^{r x} If the roots are complex and equal to a \pm b i , by Euler's formula the solution simplifies to : y=e^{a x} (C_1 \cos (b x) + C_2 \sin (b x)) The constant of integration is not restricted to real numbers in this case. Method of undetermined coefficients This method is used to find a particular solution for non-homogeneous equations by using a similar function as a potential particular solution YP and adjusting the coefficients. It is especially useful when the forcing term is a polynomial, trigonometric function, or exponential function, however if there is information known about the particular integral required different guesses can be tried. For example: : y'' + 3y' - 4y = 3x^2 : Y_p = Ax^2 + Bx + C, \;\; Y'_p = 2Ax + B, \;\; Y''_p = 2A : 2A + 3(2Ax + B) - 4(Ax^2 + Bx + C) = 3x^2 This equality now creates a linear system of equations. : -4A = 3 : 6A - 4B = 0 : 2A + 3B - 4C = 0 : A = -\frac{3}{4}, \;\; B = -\frac{9}{8}, \;\; C = -\frac{39}{32} : Y_p = -\frac{3}{4} x^2 -\frac{9}{8} x -\frac{39}{32} This particular solution, combined with the general solution of the associated homogeneous equation, gives a final general solution of : y = C_1 e^{-4x} + C_2 e^{x} -\frac{3}{4} x^2 -\frac{9}{8} x -\frac{39}{32} Real world examples Murder case Detectives find a murder victim in room with the thermostat set to 20 °C at 4:30 AM. The detectives take a measurement as soon as they arrive and find the body to be 27 °C. An hour later the body is at 25 °C. What time was the victim killed? This problem can be solved by using Newton's Law of Cooling, which states : \frac{dT}{dt}=-k(T-a) T is temperature, the dependent variable, t is time, the independent variable, a is the ambient temperature and k is a constant. Since the variables can be separated, we can find the solution using that method. : \frac{dT}{dt}=-k(T-a) : \frac{dT}{(T-a)}=-kdt : \int \frac{dT}{(T-a)}=\int -kdt : \ln(T-a)=-kt + C : T=Ce^{-kt} + a : T=Ce^{-kt} + 20 Firstly, we must find C . We can do this with the initial value T(0) = 36.8 (human body temperature). : 36.8=Ce^0 + 20 : 36.8=C + 20 : C=16.8 We can now solve for x and k . We will assume that 4:30 is x hours after death. : 27=16.8e^{-kx} + 20, 25=16.8e^{-k(x+1)} + 20 : k=-\frac{\ln(\frac{7}{16.8})}{x}=\frac{0.875}{x}, k=-\frac{\ln(\frac{5}{16.8})}{x+1} = \frac{1.212}{x+1} : \frac{0.875}{x} = \frac{1.212}{x+1} : x = 2.597 The murder took place 2.597 hours ago, or at 1:56 AM. We can also find k and find our final solution, although this is unnecessary as far as the detectives are concerned since we already know the time of death. : 27=16.8e^{-2.597k} + 20 : k=\frac{0.875}{x}=\frac{0.875}{2.597}=0.337 : T=16.8e^{-0.337t} + 20 Radioactive decay Polunium-208 has a half life of 2.898 years. If the original sample was 10 grams, how much will remain after 1 year? We can set this up as a differential equation, with A as the amount, t as time, and k as a constant. : \frac{dA}{dt} = kA This is a separable differential equation, and can be solved to give : \ln(A) = kt + C : A = Ce^{kt} Since we have the initial value A(0) = 10 , we can find C. : 10 = Ce^{k0} = Ce^0 = C : A = 10e^{kt} k can be found since we know that after 2.898 years there will be 5 grams. : 5 = 10e^{2.898k} : k = \frac{\ln(\frac{1}{2})}{2.898} = -0.239 We know have a complete formula, which we can use to calculate out answer. : A = 10e^{-0.239t} : A(1) = 10e^{-0.239} = 10(0.787) = 7.87 Our amount of Polunium-208 after 1 year is 7.87 grams. Simple harmonic motion A system which obeys Hooke's law, or : F = -kx can be rewritten as a second-order differential equation. Using Newton's 2nd Law, : \sum F = ma : \implies F = - kx = m \frac{d^2 x}{dt^2} : \frac{d^2 x}{dt^2} + \frac{k}{m} x = 0 This equation will have the characteristic equation : r^2 + \frac{k}{m} = 0 : r = \pm \sqrt{-\tfrac{k}{m}} = \pm \sqrt{\tfrac{k}{m}} i = \pm \omega i The solution to the differential equation is : x(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} = C_1 e^{\omega i t} + C_2 e^{-\omega i t} By using Euler's formula this can be written in the form : x(t) = C_1 (\cos (\omega t) + i \sin(\omega t)) + C_2 (\cos (\omega t) - i \sin(\omega t)) : x(t) = (C_1 + C_2) \cos (\omega t) + (C_1 - C_2) \sin (\omega t) : x(t) = C_3 \cos (\omega t) + C_4 \sin (\omega t) which, by trigonometric identities, can be written as : x(t) = A \cos (\omega t - \varphi) where A = \sqrt{C_3^{ \ 2} + C_4^{ \ 2}} and \tan \varphi = \tfrac{C_4}{C_3} . Systems which follow this equation are said to exhibit simple harmonic motion. Category:Analysis Category:Ordinary differential equations